Is ${62784}$ divisible by $9$ ?
A number is divisible by $9$ if the sum of its digits is divisible by $9$ . [ Why? First, we can break the number up by place value: $ \begin{eqnarray} {62784}= &&{6}\cdot10000+ \\&&{2}\cdot1000+ \\&&{7}\cdot100+ \\&&{8}\cdot10+ \\&&{4}\cdot1 \end{eqnarray} $ Next, we can rewrite each of the place values as $1$ plus a bunch of $9$ s: $ \begin{eqnarray} {62784}= &&{6}(9999+1)+ \\&&{2}(999+1)+ \\&&{7}(99+1)+ \\&&{8}(9+1)+ \\&&{4} \end{eqnarray} $ Now if we distribute and rearrange, we get this: $ \begin{eqnarray} {62784}= &&\gray{6\cdot9999}+ \\&&\gray{2\cdot999}+ \\&&\gray{7\cdot99}+ \\&&\gray{8\cdot9}+ \\&& {6}+{2}+{7}+{8}+{4} \end{eqnarray} $ Any number consisting only of $9$ s is a multiple of $9$ , so the first four terms must all be multiples of $9$ That means that to figure out whether the original number is divisible by $9 $ , all we need to do is add up the digits and see if the sum is divisible by $9$ . In other words, ${62784}$ is divisible by $9$ if ${ 6}+{2}+{7}+{8}+{4}$ is divisible by $9$ Add the digits of ${62784}$ $ {6}+{2}+{7}+{8}+{4} = {27} $ If ${27}$ is divisible by $9$ , then ${62784}$ must also be divisible by $9$ ${27}$ is divisible by $9$, therefore ${62784}$ must also be divisible by $9$.